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Question Number 205727 by mathzup last updated on 28/Mar/24

calculate lim_(x→0) ((e^x −cosx)/x^2 )

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{e}^{{x}} −{cosx}}{{x}^{\mathrm{2}} } \\ $$

Commented by lepuissantcedricjunior last updated on 29/Mar/24

lim_(x→0) ((e^x −cosx)/x^2 )=(0/0)=FI ⇔lim_(x→0) ((e^x −1−(cosx−1))/x^2 ) ⇔lim_(x→0) (((e^x −1)/x^2 ) −((cosx−1)/x^2 )) dl e^x =1+x+(x^2 /(2!))+(x^3 /(3!))+x^3 (0) dl cos(x)=1−(x^2 /(2!))+(x^4 /(4!))+x^4 (0) lim_(x→0) ((1+x+(x^2 /2)+(x^3 /(3!))−1−1+(x^2 /2)−(x^4 /4)+1)/x^2 ) ⇔lim_(x→0) ((1+x+(x^2 /(3!))−(x^3 /(4!)))/x)=(1/0)=∞ lim_(x→0) ((e^x −cosx)/x^2 )=∞ .............le puissant Dr......................

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\boldsymbol{{cosx}}}{\boldsymbol{{x}}^{\mathrm{2}} }=\frac{\mathrm{0}}{\mathrm{0}}=\boldsymbol{\mathrm{FI}} \\ $$$$\Leftrightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}−\left(\boldsymbol{{cosx}}−\mathrm{1}\right)}{\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }\:−\frac{\boldsymbol{{cosx}}−\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }\right) \\ $$$$\boldsymbol{{dl}}\:\boldsymbol{{e}}^{\boldsymbol{{x}}} =\mathrm{1}+\boldsymbol{{x}}+\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{3}!}+\boldsymbol{{x}}^{\mathrm{3}} \left(\mathrm{0}\right) \\ $$$$\boldsymbol{{dl}}\:\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)=\mathrm{1}−\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\boldsymbol{{x}}^{\mathrm{4}} }{\mathrm{4}!}+\boldsymbol{{x}}^{\mathrm{4}} \left(\mathrm{0}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\boldsymbol{{x}}+\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{2}}+\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{3}!}−\mathrm{1}−\mathrm{1}+\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{2}}−\frac{\boldsymbol{{x}}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\boldsymbol{{x}}+\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{3}!}−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{4}!}}{\boldsymbol{{x}}}=\frac{\mathrm{1}}{\mathrm{0}}=\infty\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\boldsymbol{{cosx}}}{\boldsymbol{{x}}^{\mathrm{2}} }=\infty \\ $$$$.............{le}\:{puissant}\:{Dr}...................... \\ $$

Commented by mr W last updated on 28/Mar/24

please post your answer to a question as “answer”, not as “comment”! thanks!

$${please}\:{post}\:{your}\:{answer}\:{to}\:{a}\:{question} \\ $$$${as}\:``{answer}'',\:{not}\:{as}\:``{comment}''!\:{thanks}! \\ $$

Commented by mr W last updated on 28/Mar/24

veuillez poster votre réponse à une question comme "réponse", pas comme "commentaire"! Merci!

Commented by York12 last updated on 29/Mar/24

Tu es Francis!?

$$ \\ $$Tu es Francis!?

Answered by mr W last updated on 28/Mar/24

=lim_(x→0) ((e^x +sin x)/(2x))=(1/0)=∞

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} +\mathrm{sin}\:{x}}{\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{0}}=\infty \\ $$

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